Problem: Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} \sin(y) & x\cos(y) \\ \\ -y\sin(x) & \cos(x) \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Answer: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} \sin(y) & x\cos(y) \\ \\ -y\sin(x) & \cos(x) \end{bmatrix} \right) \\ \\ &= \cos(x)\sin(y) + xy\cos(y)\sin(x) \end{aligned}$ If we evaluate $|J(f)|$ at $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$, we get $0$. Because the Jacobian determinant here is equal to $0$, we can conclude that $f$ will infinitely contract the space around $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$. To recap, the Jacobian determinant of $f$ is $\cos(x)\sin(y) + xy\cos(y)\sin(x)$, and $f$ will infinitely contract the space around the point $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$.